## Question 13(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{A} = \begin{bmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta\end{bmatrix}$ where $\tan\theta = m$ | B1 | States $m = \tan\theta$; condone omission of $\mathbf{A}=$ |
| $\mathbf{A} = \begin{bmatrix}\cos^2\theta - \sin^2\theta & 2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \sin^2\theta - \cos^2\theta\end{bmatrix}$ | M1 | Replaces $\sin 2\theta$ and $\cos 2\theta$ with expressions in trig ratios of $\theta$ or in terms of $m$ |
| $= \cos^2\theta\begin{bmatrix}1-\tan^2\theta & 2\tan\theta \\ 2\tan\theta & \tan^2\theta - 1\end{bmatrix} = \frac{1}{\sec^2\theta}\begin{bmatrix}1-m^2 & 2m \\ 2m & m^2-1\end{bmatrix}$ | M1 | Deduces one element can be expressed as $\cos^2\theta(1-\tan^2\theta)$ or $\cos^2\theta(1-m^2)$ or $2\tan\theta(\cos^2\theta)$ or $2m\cos^2\theta$; or uses $\cos 2\theta = \frac{1-m^2}{1+m^2}$ or $\sin 2\theta = \frac{2m}{1+m^2}$ |
| Uses $\cos^2\theta = \frac{1}{m^2+1}$ | M1 | |
| $\mathbf{A} = \left(\frac{1}{m^2+1}\right)\begin{bmatrix}1-m^2 & 2m \\ 2m & m^2-1\end{bmatrix}$ | R1 | Completes reasoned argument; must include $\mathbf{A}=$ |

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## Question 13(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{BA} = \left(\frac{1}{m^2+1}\right)\begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}1-m^2 & 2m \\ 2m & m^2-1\end{bmatrix} = \left(\frac{1}{m^2+1}\right)\begin{bmatrix}3(1-m^2) & 6m \\ 6m & 3(m^2-1)\end{bmatrix}$ | M1 | Obtains $\mathbf{BA}$ (condone $\mathbf{AB}$) or uses $\mathbf{B}=3\mathbf{I}$ |
| $(\mathbf{BA})^2 = \left(\frac{1}{m^2+1}\right)^2\begin{bmatrix}3(1-m^2) & 6m \\ 6m & 3(m^2-1)\end{bmatrix}\begin{bmatrix}3(1-m^2) & 6m \\ 6m & 3(m^2-1)\end{bmatrix}$ | M1 | Squares their $\mathbf{BA}$ or $\mathbf{AB}$ or $3\mathbf{IA}$ and simplifies |
| $= \left(\frac{1}{m^2+1}\right)^2\begin{bmatrix}9(1-m^2)^2+36m^2 & 0 \\ 0 & 36m^2+9(m^2-1)^2\end{bmatrix}$ | | |
| $= \left(\frac{1}{m^2+1}\right)^2\begin{bmatrix}9+18m^2+9m^4 & 0 \\ 0 & 9+18m^2+9m^4\end{bmatrix} = \begin{bmatrix}9 & 0 \\ 0 & 9\end{bmatrix} = 9\mathbf{I}$ | R1 | Completes reasoned argument using $\mathbf{BA}$ not $\mathbf{AB}$; or uses and states $A^2=\mathbf{I}$ because it is a reflection; condone missing $(\mathbf{BA})^2=$ |

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## Question 13(c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses fact that $\mathbf{B}$ represents an enlargement scale factor 3, or that $\mathbf{A}$ represents a reflection in $y=mx$ | M1 | |
| Draws four correct lines on diagram (arrows and dashed line not needed), labels $P'$; condone $\mathbf{A}$ and $\mathbf{B}$ reversed | A1 | |

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## Question 13(c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| The lines on the diagram show the effect of $\mathbf{A}$, then $\mathbf{B}$, then $\mathbf{A}$ again, then $\mathbf{B}$ again, on point $P$ | E1 | Explains how at least one of their lines represents a relevant transformation |
| The end point is the result of transforming $P$ by an enlargement of scale factor 9, centre of enlargement $O$ | E1 | Explains how their lines represent the result of the transformations |

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## Question 13(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{C} = \mathbf{BA}$; $\begin{bmatrix}\frac{12}{5} & \frac{9}{5} \\ \frac{9}{5} & -\frac{12}{5}\end{bmatrix} = \left(\frac{1}{m^2+1}\right)\begin{bmatrix}3(1-m^2) & 6m \\ 6m & 3(m^2-1)\end{bmatrix}$ | B1 | Expresses $\mathbf{BA}$ correctly in terms of $m$, or calculates $\mathbf{B}^{-1}\mathbf{C}$ correctly |
| $\frac{9}{5} = \frac{6m}{m^2+1}$, so $3m^2 - 10m + 3 = 0$ | M1 | Sets up equation in $m$ |
| $m = 3$ or $m = \frac{1}{3}$ | A1 | Finds a correct solution; condone other solution(s) not rejected |
| $m=3$ gives $\frac{3(1-m^2)}{m^2+1} = -\frac{12}{5} \neq \frac{12}{5}$, so discard $m=3$; correct value is $m = \frac{1}{3}$ | R1 | Uses rigorous argument to obtain required result, including clear reason for choosing $m=\frac{1}{3}$ |

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